What is the tangent of 1/(2sqrt(x)) ?

The tangent of 1/(2sqrt(x)) is y=-1/(4a_{0)^{3/2}}x+3/(4sqrt(a_{0))}

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tangent of 1/(2sqrt(x))

tangent of

\frac{1}{2 x}

y = - \frac{1}{4 a _{0}^{\frac{3}{2}}} x + \frac{3}{4 a _{0}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{2 a _{0}})

Find the slope of

y = \frac{1}{2 x} : \frac{d y}{d x} = - \frac{1}{4 x ^{\frac{3}{2}}}

Compute the slope of

y = \frac{1}{2 x}

x = a_{0} : m = - \frac{1}{4 a _{0}^{\frac{3}{2}}}

Find the line with slope m=

- \frac{1}{4 a _{0}^{\frac{3}{2}}}

and passing through

(a_{0}, \frac{1}{2 a _{0}}) : y = - \frac{1}{4 a _{0}^{\frac{3}{2}}} x + \frac{3}{4 a _{0}}

y = - \frac{1}{4 a _{0}^{\frac{3}{2}}} x + \frac{3}{4 a _{0}}

\frac{d}{d x} (arcsin (1 - x^{2}))

\int \frac{( x ^{3} - 1 )}{x ^{2} ( x - 2 ) ^{3}} d x

\frac{\partial}{\partial x} (e^{1 + y - x})

\int 3 x - \frac{2}{x} d x

y^{^{'}} = y - t, y (0) = 0.5

What is the tangent of 1/(2sqrt(x)) ?
The tangent of 1/(2sqrt(x)) is y=-1/(4a_{0)^{3/2}}x+3/(4sqrt(a_{0))}